Question

Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol. ATP + H2O → ADP + Pi Which expression gives the ratio of ADP to ATP at equilibrium, if the [Pi] = 1.0 M? (Note: Use RT = 2.5 kJ/mol.)

Answer 1

`6.14\cdot 10^(-6)`

Step-by-step explanation:

Firstly, write the expression for the equilibrium constant of this reaction:

`K_(eq) = ([ADP][Pi])/(ATP)`

Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

`\Delta G^o = -RT ln K_(eq)`

From here, rearrange the equation to solve for K:

`K_(eq) = e^{-(\Delta G^o)/(RT)}`

Now we know from the initial equation that:

`K_(eq) = ([ADP][Pi])/(ATP)`

Let's express the ratio of ADP to ATP:

`([ADP])/([ATP]) = ([Pi])/(K_(eq))`

Substitute the expression for K:

`([ADP])/([ATP]) = ([Pi])/(K_(eq)) = \frac{[Pi]}{e^{-(\Delta G^o)/(RT)}}`

Now we may use the values given to solve:

`([ADP])/([ATP]) = ([Pi])/(K_(eq)) = \frac{[Pi]}{e^{-(\Delta G^o)/(RT)}} = [Pi]e^{(\Delta G^o)/(RT)} = 1.0 M\cdot e^{(-30 kJ/mol)/(2.5 kJ/mol)} = 6.14\cdot 10^(-6)`